Levente Szekrenyes

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since Mar 08, 2002
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Recent posts by Levente Szekrenyes

Thanks for the answers. It seems like there is no easy way Gonna need to rewrite a few classes.

Too bad a thread cannot be set to daemon after it was started.
The problem is, I have several background threads running, which don't stop just because I'm disposing the main window, so my application isn't "exiting", it just doesn't show anything on screen.
Hi,

I have a problem with this Swing application we develop.It uses a lot of threads running in the background, and when the user closes it, the application graciously calls System.exit. However, this applications needs to be called from within another one (for testing purposes), and when calling System.exit, the "parent" application also dies.

So my question is: how can I close the application, without calling System.exit?
Hi,
Did anyone here use JDataStore (whatever version)? I need some opinions. I have to see whether it's worth switching from a native RDBMS solution (on Windows) to Borland's product...
Thanks!
[ February 05, 2003: Message edited by: Levente Szekrenyes ]
21 years ago
Well, in your case 3 hours per day for one month *might* be enough. But I suggest you to take as many mock exams as possible. The rule of thumb is that when you score over 80% on most of them, you are ready. Six months is, in my opinion, way too much, if you have some experience with Java an are an OO programmer.
As far as I know, the voucher is valid for one year from the date of purchase, so you'll have to register for the exam during this timeframe, otherwise the voucher will expire.
This has to do with the way pre- and postincrements work. With i++ (post-increment), first i is evaluated, then it is incremented. With ++i(pre-increment), first i is incremented, then evaluated.
int a = i++ - ++i
This works as follows: Initially, i is 0. When trying to do the substraction, the compiler evaluates the first operand, then the second, then it does the substraction.
So, first operand is evaluated to 0, then i is incremented, so it becomes 1. Then, second operand is evaluated, so first i is incremented, it becomes 2, then follows the evaluation, resulting in the current value of i which is 2. So now we have first operand = 0, second operand = 2, substraction results in -2.

Originally posted by Alfred Kemety:

Levente, the 1.4 is new, still being revised, 1.2 has been there for quite a while... I think...


Is the 1.4 still being revised? I mean, is it possible that the questions might change?
There was a beta-testing phase for 1.4 (Sun even offered free vouchers for that), and I thought that after this phase the exam was final and nothing will ever change about it (question wording, format etc)
But this is really off-topic, so perhaps we should open another topic...or just leave it to the bartenders to move these posts somewhere else

[ December 12, 2002: Message edited by: Levente Szekrenyes ]
Ben,
You took the 1.4 exam, while I took 1.2 . Should we conclude that the 1.2 version is much better designed and more precisely worded? Because I really meant what I said, I didn't have any interpretation problems during the exam (and I'm not a native English speaker)
Well, Dan's right, all the variants are "possible" outputs of the program. By choosing g, you're basically saying that none of the variants given can ever occur as an output.
However, you're also right, that this question might be a little bit ambiguos. But rest assured, the real exam questions are very well tested linguistically, so you'll have no doubts there. The wording is very clear so you'll know exactly what you're asked to answer.


Tightly encapsulated classes hide their state variables from outside interference, typically marking all member variables as private and providing appropriate methods for interaction with the instances.


from the RHE book
Here you have a link : Chapter 6 : Objects and Classes, from the RHE book
Keen is right, your exception is handled correctly, that's why the text "after finally" prints out. If you modify your code like this:

you will notice that your exception is not properly caught, so when run, the program prints out:
"before exception"
"333"
and then just exits with an error messsage, without printing "after finally".
Note that, because the exception wasn't properly caught, I also had to declare that the main method might throw an Exception...

Originally posted by Vedhas Pitkar:
Hi has the exam number changed?COz when I gave the exam I clearly remember the exam no. as 310-025.


The exam no. for the 1.2 version is 310-025, for the 1.4 version it is 310-035. The Developer exam is 310-027.
Ji,
I didn't really get your question. Do you have a specific scenario in your mind (then give us some code) or you're just wondering what's the difference between compile-time and run-tim errors?
Thomas,
Does that mean that after two years, a SCJP cannot go for SCJD?